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+--- Thread: Corn's Calculus Problems (/showthread.php?tid=34091)
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Corn's Calculus Problems - Shidoshi - 2011-02-22
The first obvious thing to do would be to do a substitution: u=e^x -> du=e^x*dx which leaves you with a simple integral of cossec(1+u) which is equal to 1/sin(1+u). From there my memory of calculus doesn't help me much, but I guess integration by parts should work, maybe.
Corn's Calculus Problems - shouri - 2011-02-22
^ uh...
better substitution:
u = e^x + 1
du = e^x dx
the problem becomes:
integral: csc u du
= ln ( csc u + cot u ) +C <~ this integral is usually given to you in your book.
proof here: http://answers.yahoo.com/question/index?qid=20080225213623AAUQMz6
(first part only)
Now substite u = e^x +1 back in.
= ln ( csc(e^x +1) + cot(e^x +1) +C
Corn's Calculus Problems - Shidoshi - 2011-02-22
Integrals in the form of INT(f(ax+b)) are always easy to solve if you know the primitive of f(x). It's just F(ax+b)/a + C (where F(x) is the primitive).
So the substitution is pretty pointless aside from making it look prettier.
Corn's Calculus Problems - shouri - 2011-02-22
yes, but letting u = ax+b will show you WHY that's the case. And knowing the WHY is much better than just knowing.
Corn's Calculus Problems - Shidoshi - 2011-02-22
I know it because I've tested it myself before. Then upon observation I just do that step mentally to make things quicker.
Corn's Calculus Problems - shouri - 2011-02-22
then next time i solve a problem for YOU, i'll do that for you. How the heck am I supposed to know if everyone knows how to take that mental shortcut? Since I don't, I'll show em the long way.